Explicit description of small open set containing the rationals We know that the set $\mathbb{Q}$ of rational numbers has measure zero because it is countable. In fact, if $(q_n)_{n=1,2,\ldots}$ is an enumeration of $\mathbb{Q}$, then $\bigcup_{n=1}^\infty(q_n-2^{-n}\varepsilon,q_n+2^{-n}\varepsilon)$ covers $\mathbb{Q}$, and this open set is a union of open intervals whose lengths add up to $2\varepsilon$.
That is nice and explicit – but those intervals will overlap (a lot, probably), and we know of course that an open set is a disjoint union of open intervals. It is not too hard to show that the combined lenghts of those intervals (the components of the open set) is less than $2\varepsilon$. However, this is far less explicit.

Question: Can you give an explicit list of pairwise disjoint open intervals of finite combined length whose union contains all rational numbers?

It is not hard to give a procedure for creating such a list, but that is not explicit enough. I would like a nice formula for the end points of the $n$th interval. (Those end points would have to be irrational, of course.)
 A: Here is a not so satisfactory answer, since there is a single parameter $\alpha$ that I can't make explicit (I just know it can be any element of a dense set, but don't know what it could be...).
The heart of the idea is to use the complement of a fat Cantor set, which has a nice expression as a union of disjoint intervals. More precisely, I adapt a construction of Boes, Darst and Erdös.
Let us create a fat Cantor set in $[\sqrt{2}, \sqrt{2}+1]$ (here $x_0 = \sqrt{2}$ is simply chosen because it is irrational, you could choose any other irrational number) with the following procedure, for a parameter $\alpha \in ]0,1[$:

*

*remove a central (open) interval in $[x_0, x_0+1]$ with length $\frac{a}{3}$, which creates two segments, from which you remove two central intervals with length $\frac{a}{3^2}$, creating four intervals, ...


*at step $n$, from the $2^n$ intervals with equal lengths, remove central intervals with lengths $\frac{a}{3^{n+1}}$.

The set which remains is a Cantor set that we will denote $C_{\alpha}$. Before choosing a proper $a$, let us specify the form of the open complement, which will (hopefully) contain all rationals in $[x_0, x_0+1]$. At step $n$, all $2^n$ segments not removed have length $l_n$, which satisfy $l_0=1$ and $l_n = \frac{l_{n-1} - \alpha 3^{-n}}{2}$, so $l_n  = 2^{-n}(1 - \frac{\alpha}{3}\sum \limits_{i=0}^{n-1} (\frac{2}{3})^i\big)$ i.e. $l_n = 2^{-n} (1-\alpha) + \alpha 3^{-n}$. Since at step $N$, the removed intervals are $$\bigcup \limits_{\varepsilon_0, ..., \varepsilon_{N-1}=0,1} \big]x_0 + \sum \limits_{n=0}^{N-1} \varepsilon_n \big(l_n + \frac{\alpha}{3^n}\big) + l_N, \ x_0 + \sum \limits_{n=0}^{N-1} \varepsilon_n \big(l_n + \frac{\alpha}{3^n}\big) + l_N + \frac{\alpha}{3^n}\big[$$ we deduce that the complement of $C_{\alpha}$ is $$[x_0, x_0+1] \backslash C_{\alpha} = \bigcup \limits_{\substack{N \in \mathbb{N} \\ (\varepsilon_n^N) \in \{0,1\}^N \\ \varepsilon_N^N=1}} \Big]x_0 + \sum \limits_{n=0}^N \varepsilon_n \big(\frac{1-\alpha}{2^n} + \frac{2\alpha}{3^n}\big) - \frac{\alpha}{3^N}, \ x_0 + \sum \limits_{n=0}^N \varepsilon_n \big(\frac{1-\alpha}{2^n} + \frac{2\alpha}{3^n}\big)\Big[$$
The procedure mentioned first makes it "obvious" that these intervals are disjoint, so I won't delve into a proof.
The second remark before choosing the proper $\alpha$ is that $\mu(C_{\alpha}) = \lim \limits_{n \to \infty} 2^n l_n = 1-\alpha$.

Now to prove that $\alpha$ can be chosen so that $C_{\alpha}$ contains no rational, Boes, Darst and Erdös write $C_{\alpha}$ as an image of $\{0,1\}^{\mathbb{N}}$ under the function $\phi_{\alpha}: S \mapsto \sum \limits_{n \in S} \frac{1-\alpha}{2^n} + \frac{2\alpha}{3^n}$. They note that:

*

*$||\phi_{\alpha}-\phi_{\beta}||_{\infty} = \frac{|\alpha-\beta|}{6}$, so for any $x$, $[x] := \{\alpha \in ]0,1[: x \in C_{\alpha}\}$ is closed


*if $\alpha \neq \beta$ and $S \notin \{\emptyset, \mathbb{N}\}$, $\phi_{\alpha}(S) \neq \phi_{\beta}(S)$


*the two previous properties imply that for $x \in ]x_0, x_0+1[$, $[x]$ is always nowhere dense: if $\phi_{\alpha}(E)=x=\phi_{\beta}(F)$ (with $\alpha \neq \beta$ and thus $E \neq F$), there exists $\gamma \in ]\alpha, \beta[$ such that $x \notin \phi_{\gamma}(\mathcal{P}(\mathbb{N})) = C_{\gamma}$
That last point is obviously crucial: if we don't want a rational $q \in ]\sqrt{2}, \sqrt{2}+1[$ in $C_{\alpha}$, we know we can choose $\alpha$ in $[q]^C = ]0,1[\backslash [q]$ which is an open dense set.
Thus the set of $\alpha$ such that $C_{\alpha}$ contains no rational is $\bigcap \limits_{q \in \mathbb{Q} \cap [\sqrt{2}, \sqrt{2}+1]} [q]^C$ which is the intersection of open dense sets. By Baire's theorem, that set is non-empty, and even dense.
By choosing $\alpha$ in that set arbitrarily close to zero, the open sets $C_{\alpha}^C$ made explicit earlier, with measure $\alpha$ arbitrarily small, contain all rational in $[\sqrt{2}, \sqrt{2}+1]$.
$ $
Even further, if you want to cover the whole real line and not only $[\sqrt{2}, \sqrt{2}+1]$, you know that for any $q \in \mathbb{Q} \cap [\sqrt{2}, \sqrt{2}+1]$, for $k \in \mathbb{N}$, the set $\{\alpha \in ]0,1[: q \notin C_{\alpha\cdot 2^{-k}}\}$ is an open dense set, so their intersection on $q, k$ is dense.
Thus there exists $\alpha \in ]0,1[$ such that for all $k \in \mathbb{N}$, $C_{\alpha / 2^k}$ contains no rational. By translation, $m + C_{\alpha / 2^k}$ contains no rational either for $m \in \mathbb{Z}$.
Thus, choosing $\alpha / 2^{-|m|}$ in the segment $[\sqrt{2}+m, \sqrt{2}+m+1]$, the following set is an answer :
\begin{align*}
\bigcup \limits_{m \in \mathbb{Z}} \bigcup \limits_{\substack{N \in \mathbb{N} \\ (\varepsilon_n^N) \in \{0,1\}^N \\ \varepsilon_N^N=1}} \Big] & \sqrt{2}+m + \sum \limits_{n=0}^N \varepsilon_n \big(\frac{1-\alpha 2^{-|m|}}{2^n} + \frac{2\alpha 2^{-|m|}}{3^n}\big) - \frac{\alpha 2^{-|m|}}{3^N}, \\ & \sqrt{2}+m + \sum \limits_{n=0}^N \varepsilon_n \big(\frac{1-\alpha 2^{-|m|}}{2^n} + \frac{2\alpha 2^{-|m|}}{3^n}\big)\Big[\end{align*}
It has measure $\sum \limits_{m \in \mathbb{Z}} \alpha 2^{-|m|} = 3 \alpha < 3$, contains all rational numbers in $\mathbb{R}$, and is made of disjoint intervals.

Final remark: although the set of possible $\alpha$ is dense in $]0,1[$, I could not find a way to exhibit an explicit one. Maybe a Liouville number-like $\alpha$ can do the trick, although I confess I have no hope of making a progress in that direction.
